Evaluate the definite integral. $\int^{-1}_{0}\left(-16\sqrt[3]{x}\right)\,dx = $
First, use the power rule: $\begin{aligned}\int^{-1}_{0}\left(-16\sqrt[3]{x}\right)\,dx~&=~\int^{-1}_{0}\left(-16x^\frac13\right)\,dx \\&=(-12x^\frac43)\Bigg|^{-1}_{0}\end{aligned}$ Second, plug in the limits of integration: $[-12\cdot({-1})^{\frac43}]-[-12\cdot0^{\frac43}] = -12+0 = -12$. The answer: $\int^{-1}_{0}\left(-16\sqrt[3]{x}\right)\,dx~=~-12$